The goal is to tile rectangles as small as possible with the given heptomino, in this case number 4 of the 108 heptominoes. We allow the addition of copies of a rectangle. For each rectangle $a\times b$, find the smallest area larger rectangle that copies of $a\times b$ plus at least one of the given heptomino will tile.

Example with the $1\times 1$ you can tile a $2\times 6$ as follows:

Now we don't need to consider $1\times 1$ further as we have found the smallest rectangle tilable with copies of the heptomino plus copies of $1\times 1$.

I found 31 more but lots of them can be found by 'expansion rules' or pattern variations. I considered component rectangles of width 1 through 11 and length to 32 but my search was far from complete.

List of known sizes:

Width 1: Lengths 1 to 15, 18, 22

Width 2: Lengths 2 to 9, 11, 15, 21

Width 3: Lengths 4, 5, 7

Width 4: Length 5

Most of these could be tiled by hand using logic rather than trial and error.

2 Answers
2

Here is a general solution for $2\times n$ rectangles, with $n$ odd and not divisible by $7$.

$2\times1$, in $2\times12$

$2\times3$, in $4\times40$

$2\times5$, in $6\times40$

and so on.

The size is $n+1$ by $2nk+10$, where $k\equiv n^{-1} \mod 7$.

Of course this general solution is not always optimal. Here are some better solutions:

$2\times1$

in $2\times9$

$2\times3$

in $8\times10$

Here is a general solution for $2\times n$ rectangles with $n$ even.

$2\times2$, in $4\times12$

$2\times4$, in $6\times26$

$2\times6$, in $8\times82$

$2\times8$, in $10\times26$

$2\times10$, in $12\times110$
and so on.

The size is $n+2$ by $2nk+10$, where $k\equiv (\frac{n}{2})^{-1} \mod 7$.

Again this general solution is not always optimal:

$2\times6$

in $8\times34$

Here is a general solution for $3\times n$ rectangles, where $n$ is not divisible by 3.

$3\times4$, in $8\times31$

$3\times5$, in $20\times38$

$3\times7$, in $14\times10$
and so on.

The size is $2nr$ by $lcm(7,n)+3$, where $r\equiv n^{-1} \mod 3$.

There is in fact a general solution for any $a\times b$ where $\gcd(a,b)=1$.

As $a$ and $b$ are co-prime, we can find positive $r$,$s$ such that $ra-sb=1$. This is equivalent to $(a-s)b-(b-r)a=1$. Doubling these equations, we can also find $t$,$u$ such that $ta-ub=2$ and $(a-u)b-(b-t)a=2$. The smallest $t$,$u$ are actually $t=(2r)\%b$ and $u=(2s)\%a$ where $\%$ is the modulo operator. This allows for the following solution:

$\begingroup$1x2=2x9, 2x2=4x12, 2x3=8x10, 2x4=6x26, 2x5=6x40, 2x6=8x34, 2x8=10x26 all minimal. 2x7 has a much smaller one than your generalisation, but 2x9=10x82, 2x11=12x54 and 2x15=16x40 are minimal (I think your formula gives those...). There is a nice smaller one for 2x21. Your 3x7=10x14 is minimal but 3x4 and 3x5 are not. Your axb generalisation seems to be minimal for 4x5=21x35.$\endgroup$
– theonetruepathJun 24 '18 at 1:46